**Answer:** because that those who don’t recognize what factorial is, 100! = 100 * 99 * 98 * … * 2 * 1

Ok, let’s look at how trailing zeros are formed in the first place. A rolling zero is developed when a lot of of 5 is multiplied v a multiple of 2. Now all we need to do is count the variety of 5’s and 2’s in the multiplication.

You are watching: How many zeros in 100!

Let’s counting the 5’s first. 5, 10, 15, 20, 25 and also so on do a complete of 20. But there is an ext to this. Due to the fact that 25, 50, 75 and 100 have two 5’s in each of castle (25 = 5 * 5, 50 = 2 * 5 * 5, …), you have to count lock twice. This makes the grand total 24. For world who like to look in ~ it native a formula suggest of view

Number of 5’s = 100/5 + 100/25 + 100/125 + … = 24 (Integer values only)

Moving on to count the variety of 2’s. 2, 4, 6, 8, 10 and so on. Full of 50 multiples the 2’s, 25 multiples the 4’s (count these once more), 12 multiples the 8’s (count these once more) and so on… The grand total comes the end to

Number of 2’s = 100/2 + 100/4 + 100/8 + 100/16 + 100/32 + 100/64 + 100/128 + … = 97 (Integer worths only)

Each pair that 2 and 5 will cause a trailing zero. Because we have only 24 5’s, we can only make 24 bag of 2’s and 5’s thus the variety of trailing zeros in 100 factorial is **24**.

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